integral from 1 to 2 of sqrt(1+9x)

Lösung

\int_{1}^{2} 1 + 9 x d x

\frac{38 19 - 20 10}{27}

Dezimale

3.79231 \dots

Schritte zur Lösung

\int_{1}^{2} 1 + 9 x d x

Wende U-Substitution an

= \int_{10}^{19} \frac{u}{9} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int_{10}^{19} u d u

Wende die Potenzregel an

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{10}^{19}

Berechne die Grenzen:

\frac{38 19 - 20 10}{3}

= \frac{1}{9} \cdot \frac{38 19 - 20 10}{3}

Vereinfache

= \frac{38 19 - 20 10}{27}

\int_{0}^{1} e^{- (2 x - 1)^{2}} d x

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